Question

Useful physical constants: g = 9.80 m/s2

1. At a baseball game in a large stadium with seats surrounding the outfield, a batter hits a “home run” up into the seats. The ball lands at a height above the height at which it was hit.

The ball is hit with an initial velocity of 43.0 m/s at an angle of 40.0 ̊ above the horizontal, and it takes 5.00 s to land. Ignore the effect of air resistance throughout this problem.

a. Decompose the ball’s initial velocity, v0, into its x- and y- components, v0x and v0y. Calculate the numerical value of both components. Choose the +y-direction to be upward and the +x-direction to be downrange. Show your work.

b. What is the baseball’s total horizontal displacement, from hit to landing? Show your work.

c. What is the maximum vertical displacement that the ball reaches during its trajectory? Show your work completely.

Answer 1

a.)

given, initial velocity = v0 = 43.0 m/s

If a vector A makes angle x with +x-axis, then

it's components are given by:

Ax = A*cos x

Ay = A*sin x

So, x-component of velocity(v0x) will be,

v0x = v0*cos

y-component of velocity(v0y) will be,

v0y = v0*cos

given, = 40.0 deg

then, v0x = 43.0*cos(40.0 deg)

v0x = 32.94 m/s

v0y = 43.0*sin(40.0 deg)

v0y = 27.64 m/s

b.

By applying kinematics law in horizontal direction,

Sx = ux*t + 0.5*ax*t^2

here, t = 5.00 sec.

ux = initial horizontal velocity = v0x

ax = horizontal acceleration = 0

then, Sx = v0x*t = 32.94*5.00

Sx = 164.7

c.)

By applying third kinematics law in vertical direction,

vy^2 - uy^2 = 2*ay*Sy

here,

uy = initial vertical velocity = v0y

vy = final vertical velocity = 0

ay = vertical acceleration = -g = -9.80 m/s^2

then, Sy = maximum vertical displacement = ??

Sy = (0^2 - v0y^2)/(2*-g)

Sy = (0^2 - 27.64^2)/(2*(-9.80))

Sy = 38.98 m

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